Orthogonal Frequency Division Multiplexing (OFDM) offers several advantages including, but not limited to, high spectral efficiency and improved immunity to multi-path fading. Consequently, OFDM is widely used in modern wireless communication systems for providing services including, but not limited to, digital audio broadcasting, digital video broadcasting and broadband internet access.
In order to provide communication services to users spread over large geographical regions, modern wireless communication systems use a cellular architecture. In the cellular architecture, a geographical region is divided into a number of sub-regions called cells. Each cell is served by a Base Station (BS). In other words, a BS of a cell provides communication services to Mobile Stations (MSs) inside the cell.
Further, for efficient utilization of resources, a cell is divided into a plurality of sectors. For example, a cell may be divided into three or more sectors. A BS of the cell serves each sector of the cell by using a set of directional antennas corresponding to each sector. The use of directional antennas permits re-use of frequency bands among sectors of the cell. Alternatively, each sector of the cell may be served by a dedicated BS.
However, since frequencies are re-used across sectors of a cell, there is a finite possibility of interference between MSs in adjacent sectors of the cell in spite of using directional antennas. In a scenario, reception of a signal at a BS in a sector from a first MS served by the BS may experience interference from a second MS served by a BS of an adjacent sector. For example, the first MS may transmit an OFDM tile embedded with a Pseudo Random Binary Sequence (PRBS). The OFDM tile includes one or more of one or more pilot sub-carriers and one or more data sub-carriers. Further, a PRBS may be embedded in the one or more pilot sub-carriers. However, the received OFDM tile may include interfering components from an OFDM tile transmitted by the second MS. This is illustrated by the following equations.y11=c(1)11h+c(2)11g+n11   (1)y13=c(1)13h+c(2)13g+n13   (2)y41=c(1)41h+c(2)41g+n41   (3)y43=c(1)43h+c(2)43g+n43   (4)where,    yij is the signal received by the BS at ith sub-carrier (pilot sub-carrier) and jth time symbol of the OFDM tile,    c(k)ij is a pseudo random binary value corresponding to ith sub-carrier and jth time symbol in the PRBS corresponding to sector k,    h is the channel between the first MS and the BS, which is constant throughout the OFDM tile,    g is the channel between the second MS and the BS, which is constant throughout the OFDM tile, and    ni,j is the noise in the channel at ith sub-carrier and jth time symbol.Further, the channel h may be estimated based on the signal received by the BS. An estimation of channel h, represented by hest, using linear averaging is given by:
                              h          est                =                ⁢                              (                                          ∑                ij                                                                              ⁢                                                y                  ij                                ⁢                                  c                                      (                    1                    )                                                  ⁢                ij                                      )                    /          4                                    (        5        )                                =                ⁢                  h          +                                    [                                                (                                                            ∑                      ij                                                                                                            ⁢                                                                                            c                          ⁡                                                      (                            1                            )                                                                          ij                                            ⁢                                                                        c                          ⁡                                                      (                            2                            )                                                                          ij                                                                              )                                /                4                            ]                        ⁢            g                    +                                    (                                                ∑                  ij                                                                                        ⁢                                  n                  ij                                            )                        /            4                                              (        6        )            The PRBS corresponding to the sector is generated independent of the PRBS corresponding to an adjacent sector. Further, the PRBS may include values of one or more of +1 and −1. Therefore, the estimation of channel h is given by:
                              h          est                =                  {                                                                      h                  +                  g                  +                                                            (                                                                        ∑                          ij                                                                                                                                ⁢                                                  n                          ij                                                                    )                                        /                    4                                                                                                with                  ⁢                                                                          ⁢                  probability                  ⁢                                                                          ⁢                                      1                    /                    16                                                                                                                        h                  +                                      g                    /                    2                                    +                                                            (                                                                        ∑                          ij                                                                                                                                ⁢                                                  n                          ij                                                                    )                                        /                    4                                                                                                with                  ⁢                                                                          ⁢                  probability                  ⁢                                                                          ⁢                                      4                    /                    16                                                                                                                        h                  +                                                            (                                                                        ∑                          ij                                                                                                                                ⁢                                                  n                          ij                                                                    )                                        /                    4                                                                                                with                  ⁢                                                                          ⁢                  probability                  ⁢                                                                          ⁢                                      6                    /                    16                                                                                                                        h                  -                                      g                    /                    2                                    +                                                            (                                                                        ∑                          ij                                                                                                                                ⁢                                                  n                          ij                                                                    )                                        /                    4                                                                                                with                  ⁢                                                                          ⁢                  probability                  ⁢                                                                          ⁢                                      4                    /                    16                                                                                                                        h                  -                  g                  +                                                            (                                                                        ∑                          ij                                                                                                                                ⁢                                                  n                          ij                                                                    )                                        /                    4                                                                                                with                  ⁢                                                                          ⁢                  probability                  ⁢                                                                          ⁢                                      1                    /                    16                                                                                                          (        7        )            
It can be observed that the probability of maximum interference from the second MS is 2/16. Further, in spite of linear averaging, the probability of partial interference is 8/16. Hence, the use of PRBS does not completely mitigate interference from MSs served by BSs of adjacent sectors.
Therefore, there is a need for new sequences, for embedding in pilot sub-carriers of an OFDM tile, which improve the channel estimation between MSs and BSs.
Skilled artisans will appreciate that elements in the figures are illustrated for simplicity and clarity and have not necessarily been drawn to scale. For example, the dimensions of some of the elements in the figures may be exaggerated relative to other elements to help to improve understanding of embodiments of the invention.